The minimum value of the function $f(x) = 3 \sin x-4 \cos x,x ∈ [-4π, 4π]$ is equal to

-5

The correct answer is Option (3) → -5

Given $f(x)=3\sin x-4\cos x$ on $[-4\pi,4\pi]$.

$R=\sqrt{3^2+(-4)^2}=5$.

$3\sin x-4\cos x=5\big(\cos\theta\sin x-\sin\theta\cos x\big)=5\sin(x-\theta)$,

with, $\cos\theta=\frac{3}{5}$ and $\sin\theta=\frac{4}{5}$.

Minimum value of $5\sin(x-\theta)$ is $-5$.

Minimum value = $-5$